What does it mean that the probability density function is proportional to a function?

Question

I'm studying for SOA/CAS Exam P and I have a problem that says that $X$ is a continuous and positive random variable whose probability density function is proportional to: $$\\frac{1}{(1+x)^5}$$ Where $0\\lt x \\lt \\infty$ and I need to find $E(X)$. But how do I use the information that $X$ is proportional to that function? What does it mean exactly?

Answer 1

Hello,

Let f(x) represent the pdf for the continuous random variable X. If f(x) is proportional to the function you give, then this simply means that f(x) equals a constant times this function. Therefore, we may write

f(x) = C/(1+x)⁵, for 0<x<∞

= 0, otherwise.

To complete the problem, you first need to find C. Using the fact that the integral of f(x) over the whole real line must equal 1, we have

∫(-∞ to ∞) f(x) dx = 1

∫(0 to ∞) C/(1 + x)⁵ dx = 1.

Carrying out the above integral (details omitted), we thus have

C/4 = 1,

so

C = 4.

Therefore, f(x) = 4/(1 + x)⁵ on the interval (0,∞). Now we can find E[X]. We have

E[X] = ∫(-∞ to ∞) xf(x) dx

E[X] = ∫(0 to ∞) 4x/(1 + x)⁵ dx.

I will leave you to carry out the details of the integration. (Integration by parts will be required). The result is

E[X] = 1/3.

Hope that helps! Let me know if you need any help with the integration or need me to clarify anything else.

William